Let $V$ be a normed vector space and $K$ a compact subset of $V$. Is the convex hull of $K$ given by $$\langle K \rangle = \left\{\sum_{i=1}^n t_i x_i\mid x_i\in K, t_i≥0\text{ s.t. }\sum_i t_i=1\right\}$$ again compact?


1 Answer 1


Answering the question with the counter example from the link in the math overflow question linked by Martin R:

Consider $$u_n=(\underbrace{0,...,0}_{n-1},1/n,0,...)$$ and $K=\bigcup_n \{u_n\} \cup \{0\}$ a compact subset of $\mathscr l^p(\mathbb N)$. The convex hull of $K$ is given by elements of the form: $$\sum_{n=1}^k a_n u_{n}\qquad\text{s.t.:}\quad \sum_{n=1}^k a_n≤1\qquad a_n≥0$$ So also $\sum_{n=1}^k 2^{-n}u_n$ lies in it. But this sequence converges to $\sum_{n=1}^\infty 2^{-n}u_n $ which does not lie in it.

However: From Theorem 5.35: The closed convex hull is compact in a complete normed vector space. So the convex hull of a compact set is pre-compact (or totally bounded if the original space is not complete).

For convenience we include the proof of the book, which shows the statement in the setting of completely metrisable locally convex vector spaces. More specifically one shows that for $K$ compact the convex hull $\langle K\rangle$ is completely bounded.

Let $\epsilon>0$, since $K$ is compact there is a finite covering of $K$ by balls of radius $\frac\epsilon2$, it is convenient to write this as: $$K\subseteq F+B_{\epsilon/2}(0)$$ for a finite set $F$. It then follows that: $$\langle K\rangle \subseteq \langle F\rangle +B_{\epsilon/2}(0)$$ because $B_{\epsilon/2}(0)$ is already convex. Now since $F$ is finite one has that $\langle F\rangle$ is compact and hence admits a covering by finitely many balls of radius $\frac\epsilon2$, write $\langle F\rangle = \widetilde F + B_{\epsilon/2}(0)$ for some finite set $\widetilde F$, then: $$\langle K \rangle \subseteq \langle F\rangle + B_{\epsilon/2}(0)\subseteq \widetilde F + B_{\epsilon/2}(0)+B_{\epsilon/2}(0)\subseteq \widetilde F + B_{\epsilon}(0)$$ Giving the conclusion that for any $\epsilon>0$ you may cover $\langle K\rangle$ by finitely many balls of radius $\epsilon$, whence $\langle K \rangle$ is totally bounded.

  • $\begingroup$ For reference: the counterexample is Example 5.34 from Aliprantis and Border, Infinite Dimensional Analysis: A Hitchhiker's Guide, Third edition, Springer. The referenced Theorem 5.35 is also from that book, but can also be found in many other functional analysis textbooks (e.g. Theorem 3.20 in Rudin). As pointed out by Robert Furber on a similar question on MathOverflow, this holds more generally in quasi-complete locally convex spaces. $\endgroup$ Jul 29, 2020 at 13:56
  • $\begingroup$ @JossevanDobbendeBruyn The question you linked was previously linked in a comment on this thread. It appears the comment has since been deleted. $\endgroup$
    – s.harp
    Jul 29, 2020 at 22:27
  • $\begingroup$ Yes, I inferred that from your answer. ☺ I was just trying to restore the reference for the benefit of other readers. $\endgroup$ Jul 29, 2020 at 22:36
  • $\begingroup$ Hopefully your comment will not get deleted. $\endgroup$
    – s.harp
    Jul 29, 2020 at 22:37
  • 1
    $\begingroup$ @Cris it is still true in locally convex completely metrisable spaces, I added the proof of the book to the answer. $\endgroup$
    – s.harp
    Sep 26, 2021 at 9:52

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